13. If a line \(\mathrm{L}\) is the line of intersection of the planes \(2 x+3 y+z=1\) and \(x+3 y+2 z=2\). If line \(\mathrm{L}\) makes an angle \(\alpha\) with the positive \(\mathrm{X}\)-axis, then the value of \(\sec \alpha\) is

Given equations:
\(\begin{aligned}
& 2 x+3 y+z=1 \\
& 2 x+3 y=1-z ... (i)\\
& x+3 y+2 z=2 \\
& x+3 y=2-2 z ... (ii)
\end{aligned}\)
Subtracting (ii) from (i), we get
\(\begin{aligned}
& 2 x+3 y-x-3 y=1-z-2+2 z \\
& x=-1+z \\
& z=\frac{x+1}{1} ... (iii)
\end{aligned}\)
Putting value of \(x\) in equation (ii), we get
\(\begin{aligned}
& -1+z+3 y=2-2 z \\
& 3 z=3-3 y \\
& z=\frac{y-1}{-1} ... (iv)
\end{aligned}\)
From (iii), (iv)
\(\frac{x+1}{1}=\frac{y-1}{-1}=\frac{z}{1}\)
Thus, angle between above line and \(\mathrm{X}\)-axis having Direction Ratio's \((1,0,0)\) is given as
\(\begin{aligned}
\cos \alpha & =\left|\frac{1 \cdot(1)+0+0}{\sqrt{1+1+1} \cdot \sqrt{1}}\right|=\frac{1}{\sqrt{3}} \\
\therefore \quad \sec \alpha & =\sqrt{3}
\end{aligned}\)